Upload
moira
View
76
Download
6
Embed Size (px)
DESCRIPTION
Chaper 2 ~ chaper 3 허승현. 제어시스템 설계. Contents. Chapter 2 동적 시스템 모델링 Chapter 3 제어시스템의 성능 및 안정도. 1) Laplace. 2) 부분분수 정리. 3) Solution of linear Ordinary Equation. 4) SFG. 5) 동적 시스템 모델링. 1) Routh - Huwitz. I. Chapter 2 동적 시스템 모델링. Chapter 2 동적 시스템 모델링. - PowerPoint PPT Presentation
Citation preview
Chaper 2 ~ chaper 3 허승현
1
Contents
I. Chapter 2 동적 시스템 모델링
II. Chapter 3 제어시스템의 성능 및 안정도
1) Laplace
2) 부분분수 정리
5) 동적 시스템 모델링
3) Solution of linear Ordinary Equation
4) SFG
1) Routh - Huwitz
2
I.Chapter 2 동적 시스템 모델링
3
I. Chapter 2 동적 시스템 모델링
1) Laplace Definition of Laplace transform
Given the real function f(t) that satisfied the condition
for some finite real , the Laplace transform of f(t) is defined as
=> One-sided Laplace transform
※ The response of a causal system does not precede the input.
causal = physically realizable
0( ) tf t e dt
0
( ) ( ) ( )stF s f t e dt L f t
operatorLaplacejs : 0 : ( ) 0 :
0 : ( ) :
t f t ignore
t f t defined
4
I. Chapter 2 동적 시스템 모델링
1) Laplace
( ) tf t e Ex1)
Multiplication by a constant
Sum and difference
( ) ( )L kf t kF s
1 2 1 2( ) ( ) ( ) ( )L f t f t F s F s
Important theorems of Laplace transform
0
( )
0
( )
1
t st
s t
F s e e dt
efor
s s
5
I. Chapter 2 동적 시스템 모델링
1) Laplace
Important theorems of Laplace transform(cont.) Differentiation
Integration
0
11 2
10
1 2 ( 1)
( )( ) lim ( ) ( ) (0)
( ) ( ) ( )( ) lim ( )
( ) (0) '(0) (0)
t
n nn n
n nt
n n n n
df tL sF s f t sF s f
dt
d f t df t d f tL sF s s f t s
dt dt dt
s F s s f s f f
1 2
0 0 00
1 2 10 0 0
( )( ) ( ) ( )
( )( )
n
st stt t
t t t
n n
F s e eL f d f d f t dt
s s sF s
L f d dt dt dts
6
I. Chapter 2 동적 시스템 모델링
1) Laplace Important theorems of Laplace transform(cont.)
Shift in time
Initial-value theorem
Final-value theorem: If sF(s) is analytic on the imaginary axis and in the right half of the s-plane, then
Ex)
( ) ( ) ( )
( ) :
Tss
s
L f t T u t T e F s
where u t Unit step function
0lim ( ) lim ( )t s
f t sF s
0lim ( ) lim ( )t s
f t sF s
2
0
5( )
( 2)
5lim ( ) lim ( )
2t s
F ss s s
f t sF s
7
I. Chapter 2 동적 시스템 모델링
1) Laplace
Important theorems of Laplace transform(cont.) Complex shifting
Real convolution(Complex multiplication)
In General
Complex convolution(Real multiplication)
( ) ( )tL e f t F s
1 2
1 2 1 20
1 20
1 2
( ) 0, ( ) 0, 0
( )* ( ) ( ) ( )
( ) ( )
( ) ( )
t
t
f t f t for t
L f t f t L f f t d
f t f t d
F s F s
11 2 1 2( ) ( ) ( ) ( )L F s F s f t f t
1 2 1 2( ) ( ) ( )* ( )L f t f t F s F s
Dual
8
I. Chapter 2 동적 시스템 모델링
2) 부분분수 정리 (Partial Fraction expansion)
G(s) has simple poles(cont.) Example
Sol)
5 3( )
( 1)( 2)( 3)
sG s
s s s
31 2
1 11
2 2
3 3
( )1 2 3
5 3( 1) ( ) 1
( 2)( 3)
( 2) ( ) 7
( 3) ( ) 6
1 7 6( )
1 2 3
ss
s
s
KK KG s
s s s
sK s G s
s s
K s G s
K s G s
G ss s s
9
I. Chapter 2 동적 시스템 모델링
2) 부분분수 정리 (Partial Fraction expansion)
G(s) has simple complex-conjugate poles
( )( )
( )( )
( ) ( )
( ) ( )
j j
j s j
P sG s
s j s j
K K
s j s j
K s j G s
10
I. Chapter 2 동적 시스템 모델링
2) 부분분수 정리 (Partial Fraction expansion) G(s) has simple complex-conjugate poles(cont.)
Example
2 2( ) ( )
2
2
1( )
( ) ( )
( )2 2
sin 1 01
n
Ls
Lt
j t j t t j t j tn n
tnn
u ts
e f t F s
g t e e e e ej j
e t t
2
2 2
2
2
2
( )2
( ) ( ) 1
2
1 1( )
2 ( ) ( )
n
n n
nj j
n
nj
n
G ss s
K Kwhere
s j s j
Kj
G sj s j s j
11
I. Chapter 2 동적 시스템 모델링
2) 부분분수 정리 (Partial Fraction expansion)
G(s) has multiple-order poles
1 2
1 2
1 2
1 22
( )
( )
( ) ( )( ) ( 1, 2, , )
( ) ( )( ) ( )( )
( ) ( )
where ( ) ( ) ( 1, 2, , )
1( ) ( )
( )!
n r
j j
rn r i
ss s
n r
rr
i i i
s i s s
r kr
k ir k s
Q s Q sG s where i n r
P s s s s s s s s s
KK K
s s s s s s
A A A
s s s s s s
K s s G s j n r
dA s s G s
r k ds
( 1, 2, , )is
k r
12
I. Chapter 2 동적 시스템 모델링
2) 부분분수 정리 (Partial Fraction expansion)
G(s) has multiple-order poles(cont.) Example
Sol)
3
1( )
( 1) ( 2)G s
s s s
0 32 1 22 3
0
2
03
3 0 11
2
1
3
( )2 1 ( 1) ( 1)
1
21
2
1 1( 1) ( ) 1
(3 3)! ( 2)
0
1
1 1 1 1( )
2 2( 2) 1 ( 1)
ss
K AK A AG s
s s s s s
K
K
dA s G s
ds s s
A
A
G ss s s s
13
I. Chapter 2 동적 시스템 모델링
3) Solution of linear Ordinary Equation
2
2
(0) 1( ) ( )3 2 ( ) 5 ( ),
'(0) 2s
yd y t dy ty t u t
ydt dt
Example
Sol)
2 5( ) (0) '(0) 3 ( ) 3 (0) 2 ( )s Y s sy y sY s y Y s
s
22
2 2
2
2
5 5( 3 2) ( ) 1
5 5( )
( 3 2) ( 1)( 2)
5 5 3
2 1 ( 2)
5 3( ) 5 0
2 2t t
s ss s Y s s
s s
s s s sY s
s s s s s s
s s s s
y t e e t
steady-state solutionor particular integral
transient solution or homogeneous solution0
lim ( ) lim ( )t s
y t sY s
14
I. Chapter 2 동적 시스템 모델링
3) Solution of linear Ordinary Equation
Example –cont. Transient
response
Steady-state response
Steady-state error( )sse 0
( ) ( )lim lim1 ( ) 1 ( )x x
R s sR sG s G s
Type of Control system2
1 2 1 22
1 2
(1 )(1 )......(1 )( )
(1 )(1 )....(1 )m m
ja b n n
K T s T s T s T sG s
s T s T s T s T s
15
I. Chapter 2 동적 시스템 모델링
3) Solution of linear Ordinary Equation
Example –cont.
16
Type of System
I
Err Const
Step RampParaboli
c
0 K 0 0
1 K 0 0
2 K 0 0
3 0 0 0
sse
Kp Kv Ka
1
R
K
R
K
R
K
I. Chapter 2 동적 시스템 모델링
4) SFG(Signal flow graph)
( )R s ( )E s ( )Y s)(sG
+_
)(sH
)(sR )(sE
( )G s
)(sY
1
)(sY
1
( )H s
17
I. Chapter 2 동적 시스템 모델링
4) SFG(Signal flow graph)
Can only be applied between an input node and an output node
A SFG : N forward paths
L loops
- the gain between the input node and output node
where
1
Nout k k
kin
y MM
y
1 2 3i
1k in out
i j kj k
M k y y
L L L
gain of the th forward path between and
( , , , )
(1 )mrL m m i j k
r r L
gain product of the th possible combination of
nontouching loops
k
k
the for that part of the SFG that is nontouching
with the th forward path
18
I. Chapter 2 동적 시스템 모델링
4) SFG(Signal flow graph) Ex)
is the same regardless of which input and output nodes are chosen.
2y 3y 4y 5y
12a23a
32a 43a
34a
44a
45a
24a
25a
5
1
1 12 23 34 45
2 12 24 45
3 12 25
?y
y
M a a a a
M a a a
M a a
11 23 32
21 34 43
31 24 43 32
41 44
12 23 32 44
L a a
L a a
L a a a
L a
L a a a
11 21 31 41 12
23 32 34 43 24 43 32 44 23 32 44
1 2 3 34 43 44
5 1 1 2 2 3 3 12 23 34 45 12 25 34 43 44 12 24 45
1 23 32 34 43 24 43 32 44
1 ( )
1 ( )
1, 1, 1 ( )
(1 )
1 ( )
L L L L L
a a a a a a a a a a a
a a a
y M M M a a a a a a a a a a a a
y a a a a a a a a
23 32 44a a a
1y
19
3 2 4 3 2 42
1
4 1 2 4
1
6 7 1 2 3 4 1 5 3 2
1 1
1 1 3 2 4 1 2 3 3
1 1 3 2 1 1 4 3 2 4 1 2 3 3 4 1 3 1 2 4
1
(1 )
(1 )
1
G H H G H Hy
y
y G G H
y
y y G G G G G G G H
y y
G H G H H G G G H
G H G H G H H G H H G G G H H G G H H H
I. Chapter 2 동적 시스템 모델링
4) SFG(Signal flow graph)
Ex)
2y 3y4y 5y
1 1G 4G
5G4H
1
3H
1y
2G 3G
6y 7y1H 2H
20
I. Chapter 2 동적 시스템 모델링
3) 동적 시스템 모델링 전기 시스템
Kirchhoff 의 법칙 적용
- - 0di
V L Ridt
diL Ri V
dt
( / )
( ) ( )
1 1( )
( ) ( / )
( ) 1 R L t
ELs R I s
s
E EI s
s Ls R R s s R L
Ei t e
R
21
문제 2.4
디스크 헤드의 위치 제어와 같은 기전 시스템에서 흔히 볼 수 있는 모터구동시스템 모델링
Ke : 기전력 상수 , Kt : 토크상수 , L : 전기자 인덕턴스R : 저항 , J1 : 회전자의 회전 관성
모터
구동부분회전축의 관성무시 . Ct : 등가 비틀림 점성 마찰계수 . t : 비틀림 스프링상수 , J2 : 부하의 회전 관성
22
II.Chapter 3 제어시스템의 성능 및 안정도
23
ll. Chapter 3 제어시스템의 성능 및 안정도
1) Routh - Huwitz
4 3 2
4
3
2
1
0
) 2 3 5 10 0
2 3 10
1 5 0
3 107 10 0
135 10
6.43 0 07
10 0 0
ex s s s s
s
s
s
s
s
2 roots in the right-half s-plane
roots : 1.0055 0.9331
0.7555 1.444
s j
s j
25
Special Cases When Routh’s Tabulation Terminates Prematurely① The 1st element in any one row of Routh’s tabulation is
zero, but the others are not.
2 roots in the right-half
s-plane
roots :
4 3 2
4
3
2
1
0
) 23 2 3 0
1 2 3
1 2 0
2 20 3 0
12 3 3
0 0
3 0 0
ex s s s
s
s
s
s
s
0.09057 0.902
0.4057 1.2928
s j
s j
0
0
1) Routh - Huwitz
ll. Chapter 3 제어시스템의 성능 및 안정도
26
2
roots :
( ) 4 4 0
( )8
( )coefficients of
marginally stable
s j
A s s
dA ss
ds
dA s
ds
5 4 3 2
5
4
3
2
1
1
0
) 4 8 8 7 4 0
1 8 7
4 8 4
32 8 28 46 6 0
4 448 24 4 4
624 24
( 0 0)4
8 0
4 0
ex s s s s s
s
s
s
s
s
s
s
1) Routh - Huwitz
ll. Chapter 3 제어시스템의 성능 및 안정도
Ex)
27
감사합니다… .28